Fixed points of differential equation

Question

Suppose we know that $\frac{dA}{dt}=h-A+kA^2$ for constants h,k. If we wish to find the fixed points for all h,k, surely we can just use the quadratic formula to get $A=\frac{1\pm\sqrt{1-4hk}}{2k}$ provided $hk<0.25$ and notice there are no fixed points for $hk>0.25$?

Answer 1

Yes, that is correct. The fixed points of

$\dfrac{dA}{dt} = h - A + kA^2 \tag 1$

occur where

$\dfrac{dA}{dt} = 0; \tag 2$

combining (1) and (2) yields

$kA^2 - A + h = h - A + kA^2 = 0; \tag 3$

at this juncture we invoke the quadratic formula, and write

$A = \dfrac{1 \pm \sqrt{1 - 4hk}}{2k}; \tag 4$

we see the solutions are real provided that

$4hk \le 1, \tag 5$

and that there are no real solutions when

$4hk > 1. \tag 6$

as

$hk \to 1 \tag 7$

from below, the two distinct values of $A$ approach one another, until the meet and become a single zero when $hk = 1$, at which point single zero of (1) disappears leaving no critical point of (1) whilst (6) binds.